What is the answer to (3+2sqrt(2))^{x/2}+(3-2sqrt(2))^{x/2}-34=0 ?

The answer to (3+2sqrt(2))^{x/2}+(3-2sqrt(2))^{x/2}-34=0 is x=4,x=-4

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(3+2sqrt(2))^{x/2}+(3-2sqrt(2))^{x/2}-34=0

(3 + 22)^{\frac{x}{2}} + (3 - 22)^{\frac{x}{2}} - 34 = 0

x = 4, x = - 4

Solution steps

(3 + 22)^{\frac{x}{2}} + (3 - 22)^{\frac{x}{2}} - 34 = 0

Apply exponent rules

((3 + 22)^{x})^{0.5} + ((3 + 22)^{x})^{- 0.5} - 34 = 0

Rewrite the equation with

(3 + 22)^{x} = u

u^{0.5} + (u)^{- 0.5} - 34 = 0

Solve

u^{0.5} + u^{- 0.5} - 34 = 0 : u = 577 + 408 \cdot 2^{\frac{1}{2}}, u = 577 - 408 \cdot 2^{\frac{1}{2}}

u = 577 + 408 \cdot 2^{\frac{1}{2}}, u = 577 - 408 \cdot 2^{\frac{1}{2}}

Substitute back

u = (3 + 22)^{x},

solve for

x

Solve

(3 + 22)^{x} = 577 + 408 \cdot 2^{\frac{1}{2}} : x = 4

Solve

(3 + 22)^{x} = 577 - 408 \cdot 2^{\frac{1}{2}} : x = - 4

x = 4, x = - 4

2^{x} = - 2

3^{x + 1} = 9

3^{x + 3} - 3^{x + 2} = 2

1^{x} = 2

(4^{x^{2}}) (2^{- 6 x}) = \frac{1}{16}

What is the answer to (3+2sqrt(2))^{x/2}+(3-2sqrt(2))^{x/2}-34=0 ?
The answer to (3+2sqrt(2))^{x/2}+(3-2sqrt(2))^{x/2}-34=0 is x=4,x=-4
Solve for x: (3+2sqrt(2))^{x/2}+(3-2sqrt(2))^{x/2}-34=0
The solution is x=4,x=-4